OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral 3 Exercise 15 (B)

Question 1. \( \int \frac{x-1}{(x+1)(x-2)} d x \)
Answer:
To solve this integral, we use partial fraction decomposition. We first set up the integral as:
Let \( \frac{x-1}{(x+1)(x-2)} = \frac{\mathrm{A}}{x+1} + \frac{\mathrm{B}}{x-2} \) ...(1)
Next, multiply both sides of equation (1) by \( (x+1)(x-2) \). This helps remove the denominators:
\( (x-1) = A(x-2) + B(x+1) \) ...(2)
Now, substitute specific values for \( x \) into equation (2) to find A and B.
First, substitute \( x = 2 \):
\( (2-1) = A(2-2) + B(2+1) \)
\( 1 = A(0) + B(3) \)
\( 1 = 3B \)
\( \implies B = \frac{1}{3} \)
Next, substitute \( x = -1 \):
\( (-1-1) = A(-1-2) + B(-1+1) \)
\( -2 = A(-3) + B(0) \)
\( -2 = -3A \)
\( \implies A = \frac{2}{3} \)
Substitute the values of A and B back into equation (1) to rewrite the integrand:
\( \frac{x-1}{(x+1)(x-2)} = \frac{2/3}{x+1} + \frac{1/3}{x-2} \)
Finally, integrate both terms:
\( I = \int \left( \frac{2/3}{x+1} + \frac{1/3}{x-2} \right) d x \)
\( I = \frac{2}{3}\int \frac{1}{x+1} d x + \frac{1}{3}\int \frac{1}{x-2} d x \)
\( \implies I = \frac{2}{3}\log|x+1| + \frac{1}{3}\log|x-2| + C \)
In simple words: We broke the complicated fraction into two simpler ones, found the missing numbers for them, and then integrated each simple fraction. Remember that the integral of \( \frac{1}{x} \) is \( \log|x| \).

🎯 Exam Tip: When using partial fractions, always remember to substitute values of \( x \) that make the terms zero (like \( x = 2 \) or \( x = -1 \) here) to quickly find the unknown constants A and B.

Question 2. \( \int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x \)
Answer:
We solve this integral using partial fraction decomposition. We start by writing the fraction as a sum of simpler fractions:
Let \( \frac{2 x-1}{(x-1)(x+2)(x-3)} = \frac{\mathrm{A}}{x-1} + \frac{\mathrm{B}}{x+2} + \frac{\mathrm{C}}{x-3} \) ...(1)
Next, multiply both sides of equation (1) by \( (x-1)(x+2)(x-3) \). This clears the denominators:
\( 2x-1 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) \) ...(2)
Now, we find the values of A, B, and C by substituting specific values of \( x \) into equation (2).
Substitute \( x = 1 \):
\( 2(1)-1 = A(1+2)(1-3) + B(0) + C(0) \)
\( 1 = A(3)(-2) \)
\( 1 = -6A \)
\( \implies A = -\frac{1}{6} \)
Substitute \( x = -2 \):
\( 2(-2)-1 = A(0) + B(-2-1)(-2-3) + C(0) \)
\( -5 = B(-3)(-5) \)
\( -5 = 15B \)
\( \implies B = -\frac{5}{15} = -\frac{1}{3} \)
Substitute \( x = 3 \):
\( 2(3)-1 = A(0) + B(0) + C(3-1)(3+2) \)
\( 5 = C(2)(5) \)
\( 5 = 10C \)
\( \implies C = \frac{5}{10} = \frac{1}{2} \)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{2x-1}{(x-1)(x+2)(x-3)} = \frac{-1/6}{x-1} + \frac{-1/3}{x+2} + \frac{1/2}{x-3} \)
Finally, integrate each term:
\( I = \int \left( \frac{-1/6}{x-1} + \frac{-1/3}{x+2} + \frac{1/2}{x-3} \right) d x \)
\( I = -\frac{1}{6}\int \frac{1}{x-1} d x - \frac{1}{3}\int \frac{1}{x+2} d x + \frac{1}{2}\int \frac{1}{x-3} d x \)
\( \implies I = -\frac{1}{6}\log|x-1| - \frac{1}{3}\log|x+2| + \frac{1}{2}\log|x-3| + C \)
In simple words: This problem involved breaking a fraction with three different factors in the bottom part into three simpler fractions. After finding the numerical values for A, B, and C, we just integrated each easy fraction separately. This method is called partial fraction decomposition.

🎯 Exam Tip: Always check your calculation of A, B, and C carefully by substituting values. Even a small error here will lead to a wrong final integral. Remember that \( \int \frac{1}{ax+b} dx = \frac{1}{a} \log|ax+b| \).

Question 3. \( \int \frac{x+7}{x^2+2 x-8} d x \)
Answer:
To solve this integral, we first factor the denominator and then use partial fractions.
Let \( I = \int \frac{x+7}{x^2+2 x-8} d x \)
Factor the denominator: \( x^2+2x-8 = (x-2)(x+4) \)
So, \( I = \int \frac{x+7}{(x-2)(x+4)} d x \)
Now, set up the partial fraction decomposition:
Let \( \frac{x+7}{(x-2)(x+4)} = \frac{\mathrm{A}}{x-2} + \frac{\mathrm{B}}{x+4} \) ...(1)
Multiply both sides of equation (1) by \( (x-2)(x+4) \) to eliminate the denominators:
\( x+7 = A(x+4) + B(x-2) \) ...(2)
Substitute values for \( x \) to find A and B.
Substitute \( x = 2 \) into equation (2):
\( 2+7 = A(2+4) + B(2-2) \)
\( 9 = A(6) + B(0) \)
\( 9 = 6A \)
\( \implies A = \frac{9}{6} = \frac{3}{2} \)
Substitute \( x = -4 \) into equation (2):
\( -4+7 = A(-4+4) + B(-4-2) \)
\( 3 = A(0) + B(-6) \)
\( 3 = -6B \)
\( \implies B = -\frac{3}{6} = -\frac{1}{2} \)
Now, substitute the values of A and B back into equation (1):
\( \frac{x+7}{(x-2)(x+4)} = \frac{3/2}{x-2} + \frac{-1/2}{x+4} \)
Finally, integrate each term:
\( I = \int \left( \frac{3/2}{x-2} - \frac{1/2}{x+4} \right) d x \)
\( I = \frac{3}{2}\int \frac{1}{x-2} d x - \frac{1}{2}\int \frac{1}{x+4} d x \)
\( \implies I = \frac{3}{2}\log|x-2| - \frac{1}{2}\log|x+4| + C \)
In simple words: First, we broke down the bottom part of the fraction into simpler multiplications. Then, we used a trick called partial fractions to change the complex fraction into two easier ones. Once we had these simple fractions, we could solve the integral easily using the logarithm rule.

🎯 Exam Tip: Always start by factoring the denominator of the integrand if it's a quadratic expression. This is the key first step for partial fraction decomposition.

Question 4. \( \int \frac{x d x}{x^2-3 x+2} \)
Answer:
To solve this integral, we first factor the denominator and then use partial fractions.
Let \( I = \int \frac{x d x}{x^2-3 x+2} \)
Factor the denominator: \( x^2-3x+2 = (x-1)(x-2) \)
So, \( I = \int \frac{x d x}{(x-1)(x-2)} \)
Now, set up the partial fraction decomposition:
Let \( \frac{x}{(x-1)(x-2)} = \frac{\mathrm{A}}{x-1} + \frac{\mathrm{B}}{x-2} \) ...(1)
Multiply both sides of equation (1) by \( (x-1)(x-2) \) to clear the denominators:
\( x = A(x-2) + B(x-1) \) ...(2)
Substitute values for \( x \) to find A and B.
Substitute \( x = 1 \) into equation (2):
\( 1 = A(1-2) + B(1-1) \)
\( 1 = A(-1) + B(0) \)
\( 1 = -A \)
\( \implies A = -1 \)
Substitute \( x = 2 \) into equation (2):
\( 2 = A(2-2) + B(2-1) \)
\( 2 = A(0) + B(1) \)
\( 2 = B \)
\( \implies B = 2 \)
Now, substitute the values of A and B back into equation (1):
\( \frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2} \)
Finally, integrate each term:
\( I = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) d x \)
\( I = -\int \frac{1}{x-1} d x + 2\int \frac{1}{x-2} d x \)
\( \implies I = -\log|x-1| + 2\log|x-2| + C \)
In simple words: This problem is solved by changing a complex fraction into simpler ones, which is called partial fraction method. We found the factors of the bottom part, then figured out the constants for each new fraction, and finally integrated them to get the answer.

🎯 Exam Tip: When the numerator's degree is less than the denominator's, partial fraction decomposition is an effective strategy. Always factor the denominator completely first.

Question 5. \( \int \frac{2 x+7}{x^2-x-2} d x \)
Answer:
To solve this integral, we begin by factoring the denominator and then applying partial fraction decomposition.
Let \( I = \int \frac{2 x+7}{x^2-x-2} d x \)
Factor the denominator: \( x^2-x-2 = (x+1)(x-2) \)
So, we have \( I = \int \frac{2 x+7}{(x+1)(x-2)} d x \)
Now, set up the partial fraction decomposition:
Let \( \frac{2 x+7}{(x+1)(x-2)} = \frac{\mathrm{A}}{x+1} + \frac{\mathrm{B}}{x-2} \) ...(1)
Multiply both sides of equation (1) by \( (x+1)(x-2) \) to clear the denominators:
\( 2x+7 = A(x-2) + B(x+1) \) ...(2)
Substitute values for \( x \) into equation (2) to find A and B.
Substitute \( x = 2 \):
\( 2(2)+7 = A(2-2) + B(2+1) \)
\( 4+7 = A(0) + B(3) \)
\( 11 = 3B \)
\( \implies B = \frac{11}{3} \)
Substitute \( x = -1 \):
\( 2(-1)+7 = A(-1-2) + B(-1+1) \)
\( -2+7 = A(-3) + B(0) \)
\( 5 = -3A \)
\( \implies A = -\frac{5}{3} \)
Now, substitute the values of A and B back into equation (1):
\( \frac{2x+7}{(x+1)(x-2)} = \frac{-5/3}{x+1} + \frac{11/3}{x-2} \)
Finally, integrate each term:
\( I = \int \left( -\frac{5/3}{x+1} + \frac{11/3}{x-2} \right) d x \)
\( I = -\frac{5}{3}\int \frac{1}{x+1} d x + \frac{11}{3}\int \frac{1}{x-2} d x \)
\( \implies I = -\frac{5}{3}\log|x+1| + \frac{11}{3}\log|x-2| + C \)
In simple words: This problem asks us to find the integral of a fraction. We first factored the bottom part of the fraction and then used a technique called partial fractions to split it into two simpler fractions. After finding the missing numbers (A and B), we integrated these simpler fractions to get our final answer.

🎯 Exam Tip: When dealing with integrals of rational functions, always check if the denominator can be factored. If it can, partial fraction decomposition is usually the most straightforward method.

Question 6. \( \int \frac{x+1}{x^2+4 x-5} d x \)
Answer:
To evaluate this integral, we first factor the denominator and then use partial fraction decomposition.
Let \( I = \int \frac{x+1}{x^2+4 x-5} d x \)
Factor the denominator: \( x^2+4x-5 = (x-1)(x+5) \)
So, \( I = \int \frac{(x+1) d x}{(x-1)(x+5)} \)
Now, set up the partial fraction decomposition:
Let \( \frac{x+1}{(x-1)(x+5)} = \frac{\mathrm{A}}{x-1} + \frac{\mathrm{B}}{x+5} \) ...(1)
Multiply both sides of equation (1) by \( (x-1)(x+5) \) to clear the denominators:
\( x+1 = A(x+5) + B(x-1) \) ...(2)
Substitute values for \( x \) into equation (2) to find A and B.
Substitute \( x = 1 \):
\( 1+1 = A(1+5) + B(1-1) \)
\( 2 = A(6) + B(0) \)
\( 2 = 6A \)
\( \implies A = \frac{2}{6} = \frac{1}{3} \)
Substitute \( x = -5 \):
\( -5+1 = A(-5+5) + B(-5-1) \)
\( -4 = A(0) + B(-6) \)
\( -4 = -6B \)
\( \implies B = \frac{-4}{-6} = \frac{2}{3} \)
Now, substitute the values of A and B back into equation (1):
\( \frac{x+1}{(x-1)(x+5)} = \frac{1/3}{x-1} + \frac{2/3}{x+5} \)
Finally, integrate each term:
\( I = \int \left( \frac{1/3}{x-1} + \frac{2/3}{x+5} \right) d x \)
\( I = \frac{1}{3}\int \frac{1}{x-1} d x + \frac{2}{3}\int \frac{1}{x+5} d x \)
\( \implies I = \frac{1}{3}\log|x-1| + \frac{2}{3}\log|x+5| + C \)
In simple words: We solved this problem by first breaking the bottom part of the fraction into two simple multiplication terms. Then, we used a technique called partial fractions to turn the whole fraction into a sum of two easier fractions, each with a constant on top. After finding these constants, we integrated each part separately, which is straightforward.

🎯 Exam Tip: Always double-check your factoring of the denominator. A mistake in factoring will lead to incorrect partial fractions and a wrong final answer.

Question 7. \( \int \frac{x^2+2x+8}{(x-1)(x-2)} d x \)
Answer:
For this integral, the degree of the numerator (2) is equal to the degree of the denominator (2). This means it is an improper fraction, so we must perform polynomial long division first.
Dividing \( x^2+2x+8 \) by \( (x-1)(x-2) = x^2-3x+2 \):
\( \frac{x^2+2x+8}{x^2-3x+2} = 1 + \frac{5x+6}{x^2-3x+2} \)
So, we can rewrite the integrand as:
\( \frac{x^2+2x+8}{(x-1)(x-2)} = 1 + \frac{5x+6}{(x-1)(x-2)} \)
Now, we use partial fraction decomposition for the fractional part:
Let \( \frac{5x+6}{(x-1)(x-2)} = \frac{\mathrm{A}}{x-1} + \frac{\mathrm{B}}{x-2} \) ...(1)
Multiply both sides of equation (1) by \( (x-1)(x-2) \) to clear denominators:
\( 5x+6 = A(x-2) + B(x-1) \) ...(2)
Substitute values for \( x \) to find A and B.
Substitute \( x = 1 \):
\( 5(1)+6 = A(1-2) + B(1-1) \)
\( 11 = A(-1) + B(0) \)
\( 11 = -A \)
\( \implies A = -11 \)
Substitute \( x = 2 \):
\( 5(2)+6 = A(2-2) + B(2-1) \)
\( 16 = A(0) + B(1) \)
\( 16 = B \)
\( \implies B = 16 \)
Now, substitute the values of A and B back into the expression for the integrand:
\( \frac{x^2+2x+8}{(x-1)(x-2)} = 1 + \frac{-11}{x-1} + \frac{16}{x-2} \)
Finally, integrate each term:
\( I = \int \left( 1 - \frac{11}{x-1} + \frac{16}{x-2} \right) d x \)
\( I = \int 1 d x - 11\int \frac{1}{x-1} d x + 16\int \frac{1}{x-2} d x \)
\( \implies I = x - 11\log|x-1| + 16\log|x-2| + C \)
In simple words: Since the top and bottom of the fraction had the same highest power of \( x \), we first divided them. This gave us a whole number part and a new fraction. We then broke this new fraction into two simpler ones using partial fractions, found the missing numbers, and integrated all parts separately.

🎯 Exam Tip: Always perform polynomial long division first if the degree of the numerator is greater than or equal to the degree of the denominator before attempting partial fraction decomposition.

Question 8. \( \int \frac{x^2-x-2}{1-x^2} dx \)
Answer:
For this integral, the degree of the numerator (2) is equal to the degree of the denominator (2). So, we must perform polynomial long division first.
First, rewrite the denominator as \( -(x^2-1) \). So the integrand is \( \frac{x^2-x-2}{-(x^2-1)} \).
By actual division:
\( \frac{x^2-x-2}{1-x^2} = -1 + \frac{-x-1}{1-x^2} \)
We can simplify the fraction further:
\( \frac{-x-1}{1-x^2} = \frac{-(x+1)}{-(x^2-1)} = \frac{x+1}{x^2-1} \)
Now, rewrite the integrand with the simplified fraction:
\( \frac{x^2-x-2}{1-x^2} = -1 + \frac{x+1}{x^2-1} \)
The term \( \frac{x+1}{x^2-1} \) can be simplified further:
\( \frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1} \)
So, the integrand becomes:
\( \frac{x^2-x-2}{1-x^2} = -1 + \frac{1}{x-1} \)
Now, we integrate each term:
\( \int \frac{x^2-x-2}{1-x^2} dx = \int \left( -1 + \frac{1}{x-1} \right) dx \)
\( \implies I = -x + \log|x-1| + C \)
In simple words: Since the top and bottom parts of the fraction had the same highest power, we first divided them. This gave us a simple number and another fraction. We then simplified this new fraction even more and finally integrated each part using basic rules.

🎯 Exam Tip: Always look for opportunities to simplify fractions algebraically before resorting to more complex methods like partial fractions. In this case, factoring \( x^2-1 \) allowed for a straightforward simplification.

Question 9. \( \int \frac{x^2+x+1}{(x-1)^3}dx \)
Answer:
To solve this integral, we use partial fraction decomposition for repeated factors.
Let \( I = \int \frac{x^2+x+1}{(x-1)^3} d x \)
Set up the partial fraction decomposition for the integrand:
Let \( \frac{x^2+x+1}{(x-1)^3} = \frac{\mathrm{A}}{x-1} + \frac{\mathrm{B}}{(x-1)^2} + \frac{\mathrm{C}}{(x-1)^3} \) ...(1)
Multiply both sides of equation (1) by \( (x-1)^3 \) to clear the denominators:
\( x^2+x+1 = A(x-1)^2 + B(x-1) + C \) ...(2)
Substitute values for \( x \) and compare coefficients to find A, B, and C.
Substitute \( x = 1 \) into equation (2):
\( (1)^2+(1)+1 = A(0)^2 + B(0) + C \)
\( 3 = C \)
Now, expand equation (2):
\( x^2+x+1 = A(x^2-2x+1) + B(x-1) + C \)
\( x^2+x+1 = Ax^2 - 2Ax + A + Bx - B + C \)
\( x^2+x+1 = Ax^2 + (-2A+B)x + (A-B+C) \)
Compare the coefficients of \( x^2 \):
\( 1 = A \)
\( \implies A = 1 \)
Compare the coefficients of \( x \):
\( 1 = -2A+B \)
Substitute \( A=1 \):
\( 1 = -2(1)+B \)
\( 1 = -2+B \)
\( \implies B = 3 \)
(We can also check the constant terms: \( 1 = A-B+C = 1-3+3 = 1 \). This confirms our values.)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{x^2+x+1}{(x-1)^3} = \frac{1}{x-1} + \frac{3}{(x-1)^2} + \frac{3}{(x-1)^3} \)
Finally, integrate each term:
\( I = \int \left( \frac{1}{x-1} + \frac{3}{(x-1)^2} + \frac{3}{(x-1)^3} \right) d x \)
\( I = \int \frac{1}{x-1} d x + 3\int (x-1)^{-2} d x + 3\int (x-1)^{-3} d x \)
\( I = \log|x-1| + 3\frac{(x-1)^{-2+1}}{-2+1} + 3\frac{(x-1)^{-3+1}}{-3+1} + C \)
\( I = \log|x-1| + 3\frac{(x-1)^{-1}}{-1} + 3\frac{(x-1)^{-2}}{-2} + C \)
\( \implies I = \log|x-1| - \frac{3}{x-1} - \frac{3}{2(x-1)^2} + C \)
In simple words: When the bottom of the fraction has a term like \( (x-1)^3 \), we use a special type of partial fraction called repeated factors. We break the fraction into three parts, find the numbers for each, and then integrate them. The powers on the bottom mean we use the power rule for integration.

🎯 Exam Tip: For repeated factors in the denominator, remember to include all powers from 1 up to the highest power, such as \( \frac{A}{(ax+b)} + \frac{B}{(ax+b)^2} + \frac{C}{(ax+b)^3} \). Always compare coefficients when substituting \( x \) values doesn't fully solve for all constants.

Question 10. \( \int \frac{\sin \theta \cos \theta}{\cos ^2 \theta-\cos \theta-2} d \theta \)
Answer:
To solve this integral, we use a substitution to simplify the expression, then apply partial fraction decomposition.
Let \( I = \int \frac{\sin \theta \cos \theta}{\cos ^2 \theta-\cos \theta-2} d \theta \)
Let \( t = \cos \theta \)
Then, \( d t = -\sin \theta d \theta \)
\( \implies -dt = \sin \theta d \theta \)
Substitute \( t \) and \( -dt \) into the integral:
\( I = \int \frac{t (-dt)}{t^2-t-2} = \int \frac{-t d t}{t^2-t-2} \)
Factor the denominator: \( t^2-t-2 = (t-2)(t+1) \)
So, \( I = \int \frac{-t d t}{(t-2)(t+1)} \)
Now, use partial fraction decomposition for the integrand:
Let \( \frac{-t}{(t-2)(t+1)} = \frac{\mathrm{A}}{t-2} + \frac{\mathrm{B}}{t+1} \) ...(1)
Multiply both sides of equation (1) by \( (t-2)(t+1) \) to clear the denominators:
\( -t = A(t+1) + B(t-2) \) ...(2)
Substitute values for \( t \) to find A and B.
Substitute \( t = 2 \):
\( -2 = A(2+1) + B(2-2) \)
\( -2 = A(3) + B(0) \)
\( -2 = 3A \)
\( \implies A = -\frac{2}{3} \)
Substitute \( t = -1 \):
\( -(-1) = A(-1+1) + B(-1-2) \)
\( 1 = A(0) + B(-3) \)
\( 1 = -3B \)
\( \implies B = -\frac{1}{3} \)
Now, substitute the values of A and B back into equation (1):
\( \frac{-t}{(t-2)(t+1)} = \frac{-2/3}{t-2} + \frac{-1/3}{t+1} \)
Finally, integrate each term:
\( I = \int \left( \frac{-2/3}{t-2} - \frac{1/3}{t+1} \right) d t \)
\( I = -\frac{2}{3}\int \frac{1}{t-2} d t - \frac{1}{3}\int \frac{1}{t+1} d t \)
\( I = -\frac{2}{3}\log|t-2| - \frac{1}{3}\log|t+1| + C \)
Substitute back \( t = \cos \theta \):
\( \implies I = -\frac{2}{3}\log|\cos \theta-2| - \frac{1}{3}\log|\cos \theta+1| + C \)
In simple words: We used a substitution to turn the trigonometric integral into a simpler one involving \( t \). Then we used partial fractions to break the new fraction into two parts, found the numbers, and integrated each part. Finally, we put \( \cos \theta \) back in place of \( t \) to get the answer in terms of \( \theta \).

🎯 Exam Tip: For integrals involving trigonometric functions, look for substitutions that transform the integral into a rational function of a new variable, which can then be solved using partial fractions.

Question 11. \( \int \frac{3 x-1}{(x-2)^2} d x \)
Answer:
To evaluate this integral, we use partial fraction decomposition for repeated linear factors.
Let \( I = \int \frac{3 x-1}{(x-2)^2} d x \)
Set up the partial fraction decomposition for the integrand:
Let \( \frac{3 x-1}{(x-2)^2} = \frac{\mathrm{A}}{x-2} + \frac{\mathrm{B}}{(x-2)^2} \) ...(1)
Multiply both sides of equation (1) by \( (x-2)^2 \) to clear the denominators:
\( 3x-1 = A(x-2) + B \) ...(2)
Substitute values for \( x \) and compare coefficients to find A and B.
Substitute \( x = 2 \) into equation (2):
\( 3(2)-1 = A(2-2) + B \)
\( 6-1 = A(0) + B \)
\( 5 = B \)
\( \implies B = 5 \)
Compare the coefficients of \( x \) in equation (2):
On the left side, the coefficient of \( x \) is 3.
On the right side, the coefficient of \( x \) is A.
So, \( A = 3 \)
Now, substitute the values of A and B back into equation (1):
\( \frac{3x-1}{(x-2)^2} = \frac{3}{x-2} + \frac{5}{(x-2)^2} \)
Finally, integrate each term:
\( I = \int \left( \frac{3}{x-2} + \frac{5}{(x-2)^2} \right) d x \)
\( I = 3\int \frac{1}{x-2} d x + 5\int (x-2)^{-2} d x \)
\( I = 3\log|x-2| + 5\frac{(x-2)^{-2+1}}{-2+1} + C \)
\( I = 3\log|x-2| + 5\frac{(x-2)^{-1}}{-1} + C \)
\( \implies I = 3\log|x-2| - \frac{5}{x-2} + C \)
In simple words: We used a special method called partial fractions for repeated factors, breaking the original fraction into two simpler ones. We found the numbers for these simpler fractions by comparing parts and substituting values. Then, we just integrated each part using the basic rules for logarithms and powers.

🎯 Exam Tip: For repeated linear factors in the denominator, make sure to include a term for each power from 1 up to the highest power of the factor. For example, for \( (x-2)^2 \), you need terms \( \frac{A}{x-2} \) and \( \frac{B}{(x-2)^2} \).

Question 12.
(i) \( \int \frac{x^2+x+1}{x^2(x+1)} d x \)
(ii) \( \int \frac{2}{(1-x)\left(1+x^2\right)} d x \)
Answer:
(i) To solve this integral, we use partial fraction decomposition.
Let \( I = \int \frac{x^2+x+1}{x^2(x+1)} d x \)
Set up the partial fraction decomposition:
Let \( \frac{x^2+x+1}{x^2(x+1)} = \frac{\mathrm{A}}{x} + \frac{\mathrm{B}}{x^2} + \frac{\mathrm{C}}{x+1} \) ...(1)
Multiply both sides of equation (1) by \( x^2(x+1) \) to clear the denominators:
\( x^2+x+1 = Ax(x+1) + B(x+1) + Cx^2 \) ...(2)
Substitute values for \( x \) and compare coefficients to find A, B, and C.
Substitute \( x = 0 \) into equation (2):
\( (0)^2+(0)+1 = A(0) + B(0+1) + C(0) \)
\( 1 = B(1) \)
\( \implies B = 1 \)
Substitute \( x = -1 \) into equation (2):
\( (-1)^2+(-1)+1 = A(0) + B(0) + C(-1)^2 \)
\( 1-1+1 = C(1) \)
\( 1 = C \)
\( \implies C = 1 \)
Now, compare the coefficients of \( x^2 \) in equation (2):
\( x^2+x+1 = Ax^2+Ax + Bx+B + Cx^2 \)
\( x^2+x+1 = (A+C)x^2 + (A+B)x + B \)
Comparing \( x^2 \) coefficients: \( 1 = A+C \)
Substitute \( C=1 \):
\( 1 = A+1 \)
\( \implies A = 0 \)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{x^2+x+1}{x^2(x+1)} = \frac{0}{x} + \frac{1}{x^2} + \frac{1}{x+1} \)
\( \frac{x^2+x+1}{x^2(x+1)} = \frac{1}{x^2} + \frac{1}{x+1} \)
Finally, integrate each term:
\( I = \int \left( \frac{1}{x^2} + \frac{1}{x+1} \right) d x \)
\( I = \int x^{-2} d x + \int \frac{1}{x+1} d x \)
\( I = \frac{x^{-2+1}}{-2+1} + \log|x+1| + C \)
\( \implies I = -\frac{1}{x} + \log|x+1| + C \)
(ii) To solve this integral, we use partial fraction decomposition involving an irreducible quadratic factor.
Let \( I = \int \frac{2}{(1-x)(1+x^2)} d x \)
Set up the partial fraction decomposition:
Let \( \frac{2}{(1-x)(1+x^2)} = \frac{\mathrm{A}}{1-x} + \frac{\mathrm{B}x+\mathrm{C}}{1+x^2} \) ...(1)
Multiply both sides of equation (1) by \( (1-x)(1+x^2) \) to clear the denominators:
\( 2 = A(1+x^2) + (Bx+C)(1-x) \) ...(2)
Substitute values for \( x \) and compare coefficients to find A, B, and C.
Substitute \( x = 1 \) into equation (2):
\( 2 = A(1+1^2) + (B(1)+C)(1-1) \)
\( 2 = A(2) + 0 \)
\( 2 = 2A \)
\( \implies A = 1 \)
Now, expand equation (2):
\( 2 = A + Ax^2 + Bx - Bx^2 + C - Cx \)
\( 2 = (A-B)x^2 + (B-C)x + (A+C) \)
Compare the coefficients of \( x^2 \):
\( 0 = A-B \)
Substitute \( A=1 \):
\( 0 = 1-B \)
\( \implies B = 1 \)
Compare the coefficients of \( x \):
\( 0 = B-C \)
Substitute \( B=1 \):
\( 0 = 1-C \)
\( \implies C = 1 \)
(We can also check the constant terms: \( 2 = A+C = 1+1 = 2 \). This confirms our values.)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{2}{(1-x)(1+x^2)} = \frac{1}{1-x} + \frac{x+1}{1+x^2} \)
Finally, integrate each term:
\( I = \int \left( \frac{1}{1-x} + \frac{x+1}{1+x^2} \right) d x \)
\( I = \int \frac{1}{1-x} d x + \int \frac{x}{1+x^2} d x + \int \frac{1}{1+x^2} d x \)
For \( \int \frac{1}{1-x} d x \): let \( u = 1-x \), then \( du = -dx \). So it is \( -\int \frac{1}{u} du = -\log|1-x| \).
For \( \int \frac{x}{1+x^2} d x \): let \( v = 1+x^2 \), then \( dv = 2x dx \). So it is \( \frac{1}{2}\int \frac{1}{v} dv = \frac{1}{2}\log|1+x^2| \).
For \( \int \frac{1}{1+x^2} d x \): this is a standard integral, \( \tan^{-1}x \).
So, \( I = -\log|1-x| + \frac{1}{2}\log|1+x^2| + \tan^{-1}x + C \)
(Note: The integral \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| \) and \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \) were used.)
In simple words: For part (i), we broke down the fraction using partial fractions for repeated and distinct linear factors. For part (ii), we used a similar method but with an extra step because one part of the denominator could not be factored further into simple linear terms. We then integrated each part carefully.

🎯 Exam Tip: When dealing with partial fractions, distinguish between linear factors, repeated linear factors, and irreducible quadratic factors, as each requires a different form of partial fraction decomposition.

Question 13. \( \int \frac{3 x-2}{(x+1)^2(x+3)} d x \)
Answer:
To solve this integral, we use partial fraction decomposition because the denominator has repeated factors.
Let \( I = \int \frac{3 x-2}{(x+1)^2(x+3)} d x \)
Set up the partial fraction decomposition for the integrand:
Let \( \frac{3 x-2}{(x+1)^2(x+3)} = \frac{\mathrm{A}}{x+1} + \frac{\mathrm{B}}{(x+1)^2} + \frac{\mathrm{C}}{x+3} \) ...(1)
Multiply both sides of equation (1) by \( (x+1)^2(x+3) \) to clear the denominators:
\( 3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2 \) ...(2)
Substitute values for \( x \) and compare coefficients to find A, B, and C.
Substitute \( x = -1 \) into equation (2):
\( 3(-1)-2 = A(0) + B(-1+3) + C(0) \)
\( -5 = B(2) \)
\( \implies B = -\frac{5}{2} \)
Substitute \( x = -3 \) into equation (2):
\( 3(-3)-2 = A(0) + B(0) + C(-3+1)^2 \)
\( -9-2 = C(-2)^2 \)
\( -11 = C(4) \)
\( \implies C = -\frac{11}{4} \)
Now, to find A, we can compare the coefficients of \( x^2 \) in equation (2).
Expand equation (2):
\( 3x-2 = A(x^2+4x+3) + B(x+3) + C(x^2+2x+1) \)
\( 3x-2 = Ax^2+4Ax+3A + Bx+3B + Cx^2+2Cx+C \)
\( 3x-2 = (A+C)x^2 + (4A+B+2C)x + (3A+3B+C) \)
Compare the coefficients of \( x^2 \):
\( 0 = A+C \)
Substitute \( C = -\frac{11}{4} \):
\( 0 = A - \frac{11}{4} \)
\( \implies A = \frac{11}{4} \)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{3x-2}{(x+1)^2(x+3)} = \frac{11/4}{x+1} - \frac{5/2}{(x+1)^2} - \frac{11/4}{x+3} \)
Finally, integrate each term:
\( I = \int \left( \frac{11/4}{x+1} - \frac{5/2}{(x+1)^2} - \frac{11/4}{x+3} \right) d x \)
\( I = \frac{11}{4}\int \frac{1}{x+1} d x - \frac{5}{2}\int (x+1)^{-2} d x - \frac{11}{4}\int \frac{1}{x+3} d x \)
\( I = \frac{11}{4}\log|x+1| - \frac{5}{2}\frac{(x+1)^{-2+1}}{-2+1} - \frac{11}{4}\log|x+3| + C \)
\( I = \frac{11}{4}\log|x+1| - \frac{5}{2}\frac{(x+1)^{-1}}{-1} - \frac{11}{4}\log|x+3| + C \)
\( \implies I = \frac{11}{4}\log|x+1| + \frac{5}{2(x+1)} - \frac{11}{4}\log|x+3| + C \)
In simple words: This problem uses partial fractions for a fraction with a repeated factor in its denominator. We broke the fraction into three simpler parts, found the numbers for each part using substitution and comparing powers of \( x \). Then we integrated each of these simpler parts to get the final answer.

🎯 Exam Tip: Always organize your work clearly when solving for constants A, B, and C in partial fractions. Use both substitution of root values and comparison of coefficients to efficiently find all constants.

Question 14. \( \int \frac{2 x d x}{\left(x^2+1\right)\left(x^2+2\right)} \)
Answer:
To solve this integral, we first use a substitution to simplify the integrand.
Let \( I = \int \frac{2 x d x}{\left(x^2+1\right)\left(x^2+2\right)} \)
Let \( t = x^2 \)
Then, \( d t = 2x d x \)
Substitute \( t \) and \( dt \) into the integral:
\( I = \int \frac{d t}{(t+1)(t+2)} \)
Now, we use partial fraction decomposition for the new integrand:
Let \( \frac{1}{(t+1)(t+2)} = \frac{\mathrm{A}}{t+1} + \frac{\mathrm{B}}{t+2} \) ...(1)
Multiply both sides of equation (1) by \( (t+1)(t+2) \) to clear the denominators:
\( 1 = A(t+2) + B(t+1) \) ...(2)
Substitute values for \( t \) to find A and B.
Substitute \( t = -1 \) into equation (2):
\( 1 = A(-1+2) + B(-1+1) \)
\( 1 = A(1) + B(0) \)
\( \implies A = 1 \)
Substitute \( t = -2 \) into equation (2):
\( 1 = A(-2+2) + B(-2+1) \)
\( 1 = A(0) + B(-1) \)
\( 1 = -B \)
\( \implies B = -1 \)
Now, substitute the values of A and B back into equation (1):
\( \frac{1}{(t+1)(t+2)} = \frac{1}{t+1} - \frac{1}{t+2} \)
Finally, integrate each term with respect to \( t \):
\( I = \int \left( \frac{1}{t+1} - \frac{1}{t+2} \right) d t \)
\( I = \int \frac{1}{t+1} d t - \int \frac{1}{t+2} d t \)
\( I = \log|t+1| - \log|t+2| + C \)
Using logarithm properties, \( \log a - \log b = \log (a/b) \):
\( I = \log\left|\frac{t+1}{t+2}\right| + C \)
Substitute back \( t = x^2 \):
\( \implies I = \log\left|\frac{x^2+1}{x^2+2}\right| + C \)
In simple words: We made the integral simpler by changing \( x^2 \) to \( t \). This made it a fraction that we could break into two easier parts using partial fractions. After finding the numbers for these parts, we integrated each one using logarithms and then put \( x^2 \) back to get the final answer.

🎯 Exam Tip: When you see an integral of a rational function involving \( x^2 \) where \( 2x dx \) is also present (or can be easily made so), a substitution like \( t=x^2 \) often simplifies the problem significantly before applying partial fractions.

Question 15. \( \int \frac{x^2+1}{x^2-1} d x \)
Answer:
To solve this integral, we first note that the degree of the numerator (2) is equal to the degree of the denominator (2). Therefore, we perform polynomial long division first.
Dividing \( x^2+1 \) by \( x^2-1 \):
\( \frac{x^2+1}{x^2-1} = \frac{x^2-1+2}{x^2-1} = 1 + \frac{2}{x^2-1} \)
So, \( I = \int \left( 1 + \frac{2}{x^2-1} \right) d x \)
Now, we apply partial fraction decomposition to the fractional part \( \frac{2}{x^2-1} \).
Factor the denominator: \( x^2-1 = (x-1)(x+1) \)
Set up the partial fraction decomposition:
Let \( \frac{2}{(x-1)(x+1)} = \frac{\mathrm{A}}{x-1} + \frac{\mathrm{B}}{x+1} \) ...(1)
Multiply both sides of equation (1) by \( (x-1)(x+1) \) to clear the denominators:
\( 2 = A(x+1) + B(x-1) \) ...(2)
Substitute values for \( x \) to find A and B.
Substitute \( x = 1 \) into equation (2):
\( 2 = A(1+1) + B(1-1) \)
\( 2 = A(2) + B(0) \)
\( 2 = 2A \)
\( \implies A = 1 \)
Substitute \( x = -1 \) into equation (2):
\( 2 = A(-1+1) + B(-1-1) \)
\( 2 = A(0) + B(-2) \)
\( 2 = -2B \)
\( \implies B = -1 \)
Now, substitute the values of A and B back into equation (1):
\( \frac{2}{x^2-1} = \frac{1}{x-1} - \frac{1}{x+1} \)
Substitute this back into the integral expression:
\( I = \int \left( 1 + \frac{1}{x-1} - \frac{1}{x+1} \right) d x \)
Finally, integrate each term:
\( I = \int 1 d x + \int \frac{1}{x-1} d x - \int \frac{1}{x+1} d x \)
\( I = x + \log|x-1| - \log|x+1| + C \)
Using logarithm properties, \( \log a - \log b = \log (a/b) \):
\( \implies I = x + \log\left|\frac{x-1}{x+1}\right| + C \)
In simple words: Since the top and bottom parts of the fraction had the same highest power, we first divided them to get a simple number and a new fraction. Then, we used partial fractions to break this new fraction into two easier parts. After finding the missing numbers, we integrated all three parts separately to get the final answer.

🎯 Exam Tip: For integrals where the numerator's degree is equal to or greater than the denominator's, always perform polynomial long division first. This simplifies the integrand into a polynomial plus a proper fraction, which is easier to integrate.

Question 16. \( \int \frac{x^2}{x^4+x^2-2} d x \)
Answer:
To solve this integral, we first use a substitution to simplify the expression, then apply partial fraction decomposition.
Let \( I = \int \frac{x^2}{x^4+x^2-2} d x \)
Let \( t = x^2 \)
Then the integrand becomes \( \frac{t}{t^2+t-2} \)
Factor the denominator: \( t^2+t-2 = (t-1)(t+2) \)
So, the integrand is \( \frac{t}{(t-1)(t+2)} \)
Now, set up the partial fraction decomposition for the integrand:
Let \( \frac{t}{(t-1)(t+2)} = \frac{\mathrm{A}}{t-1} + \frac{\mathrm{B}}{t+2} \) ...(1)
Multiply both sides of equation (1) by \( (t-1)(t+2) \) to clear the denominators:
\( t = A(t+2) + B(t-1) \) ...(2)
Substitute values for \( t \) to find A and B.
Substitute \( t = 1 \) into equation (2):
\( 1 = A(1+2) + B(1-1) \)
\( 1 = A(3) + B(0) \)
\( 1 = 3A \)
\( \implies A = \frac{1}{3} \)
Substitute \( t = -2 \) into equation (2):
\( -2 = A(-2+2) + B(-2-1) \)
\( -2 = A(0) + B(-3) \)
\( -2 = -3B \)
\( \implies B = \frac{2}{3} \)
Now, substitute the values of A and B back into the expression for the integrand:
\( \frac{t}{(t-1)(t+2)} = \frac{1/3}{t-1} + \frac{2/3}{t+2} \)
Substitute back \( t = x^2 \):
\( \frac{x^2}{(x^2-1)(x^2+2)} = \frac{1/3}{x^2-1} + \frac{2/3}{x^2+2} \)
Now, integrate each term with respect to \( x \):
\( I = \int \left( \frac{1/3}{x^2-1} + \frac{2/3}{x^2+2} \right) d x \)
\( I = \frac{1}{3}\int \frac{1}{x^2-1} d x + \frac{2}{3}\int \frac{1}{x^2+2} d x \)
For \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| \), use \( a=1 \).
For \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \), use \( a=\sqrt{2} \).
\( I = \frac{1}{3}\left( \frac{1}{2(1)}\log\left|\frac{x-1}{x+1}\right| \right) + \frac{2}{3}\left( \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) \right) + C \)
\( \implies I = \frac{1}{6}\log\left|\frac{x-1}{x+1}\right| + \frac{2}{3\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + C \)
We can simplify \( \frac{2}{3\sqrt{2}} \) to \( \frac{\sqrt{2}}{3} \):
\( \implies I = \frac{1}{6}\log\left|\frac{x-1}{x+1}\right| + \frac{\sqrt{2}}{3}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + C \)
In simple words: We first used a trick to replace \( x^2 \) with \( t \) to make the fraction simpler. Then we broke this new fraction into two easier parts. After finding the numbers for these parts, we put \( x^2 \) back and integrated using special formulas for \( \frac{1}{x^2-a^2} \) and \( \frac{1}{x^2+a^2} \).

🎯 Exam Tip: When the integrand contains only even powers of \( x \), a substitution like \( t=x^2 \) can convert it into a rational function of \( t \), simplifying the partial fraction decomposition. Remember to convert back and use the appropriate integral formulas for \( \frac{1}{x^2 \pm a^2} \).

Question 17. \( \int \frac{x^2+x+1}{(x+1)^2(x+2)} d x \)
Answer:
To solve this integral, we use partial fraction decomposition because the denominator has repeated and distinct linear factors.
Let \( I = \int \frac{x^2+x+1}{(x+1)^2(x+2)} d x \)
Set up the partial fraction decomposition for the integrand:
Let \( \frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{\mathrm{A}}{x+1} + \frac{\mathrm{B}}{(x+1)^2} + \frac{\mathrm{C}}{x+2} \) ...(1)
Multiply both sides of equation (1) by \( (x+1)^2(x+2) \) to clear the denominators:
\( x^2+x+1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2 \) ...(2)
Substitute values for \( x \) and compare coefficients to find A, B, and C.
Substitute \( x = -1 \) into equation (2):
\( (-1)^2+(-1)+1 = A(0) + B(-1+2) + C(0) \)
\( 1-1+1 = B(1) \)
\( 1 = B \)
\( \implies B = 1 \)
Substitute \( x = -2 \) into equation (2):
\( (-2)^2+(-2)+1 = A(0) + B(0) + C(-2+1)^2 \)
\( 4-2+1 = C(-1)^2 \)
\( 3 = C(1) \)
\( \implies C = 3 \)
To find A, we can compare the coefficients of \( x^2 \) in equation (2).
Expand equation (2):
\( x^2+x+1 = A(x^2+3x+2) + B(x+2) + C(x^2+2x+1) \)
\( x^2+x+1 = Ax^2+3Ax+2A + Bx+2B + Cx^2+2Cx+C \)
\( x^2+x+1 = (A+C)x^2 + (3A+B+2C)x + (2A+2B+C) \)
Compare the coefficients of \( x^2 \):
\( 1 = A+C \)
Substitute \( C=3 \):
\( 1 = A+3 \)
\( \implies A = -2 \)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} \)
Finally, integrate each term:
\( I = \int \left( \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} \right) d x \)
\( I = -2\int \frac{1}{x+1} d x + \int (x+1)^{-2} d x + 3\int \frac{1}{x+2} d x \)
\( I = -2\log|x+1| + \frac{(x+1)^{-2+1}}{-2+1} + 3\log|x+2| + C \)
\( I = -2\log|x+1| + \frac{(x+1)^{-1}}{-1} + 3\log|x+2| + C \)
\( \implies I = -2\log|x+1| - \frac{1}{x+1} + 3\log|x+2| + C \)
In simple words: This problem involves breaking a fraction with both a repeated factor and a different factor in the bottom part into three simpler fractions. We found the numbers (A, B, C) for these parts by using a mix of substituting values and comparing the powers of \( x \). After finding these numbers, we integrated each simple fraction using basic logarithm and power rules.

🎯 Exam Tip: When the denominator has both repeated and distinct linear factors, ensure you set up the partial fraction form correctly for each type. For \( (ax+b)^n \), include terms up to \( \frac{A_n}{(ax+b)^n} \).

Question 18. \( \int \frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)} d x \)
Answer:
To solve this integral, we first use a substitution to simplify the integrand.
Let \( I = \int \frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)} d x \)
Let \( t = x^2 \)
Then the integrand becomes \( \frac{t}{(t+1)(t+4)} \)
Now, set up the partial fraction decomposition for the new integrand:
Let \( \frac{t}{(t+1)(t+4)} = \frac{\mathrm{A}}{t+1} + \frac{\mathrm{B}}{t+4} \) ...(1)
Multiply both sides of equation (1) by \( (t+1)(t+4) \) to clear the denominators:
\( t = A(t+4) + B(t+1) \) ...(2)
Substitute values for \( t \) to find A and B.
Substitute \( t = -1 \) into equation (2):
\( -1 = A(-1+4) + B(-1+1) \)
\( -1 = A(3) + B(0) \)
\( -1 = 3A \)
\( \implies A = -\frac{1}{3} \)
Substitute \( t = -4 \) into equation (2):
\( -4 = A(-4+4) + B(-4+1) \)
\( -4 = A(0) + B(-3) \)
\( -4 = -3B \)
\( \implies B = \frac{4}{3} \)
Now, substitute the values of A and B back into equation (1):
\( \frac{t}{(t+1)(t+4)} = \frac{-1/3}{t+1} + \frac{4/3}{t+4} \)
Substitute back \( t = x^2 \):
\( \frac{x^2}{(x^2+1)(x^2+4)} = \frac{-1/3}{x^2+1} + \frac{4/3}{x^2+4} \)
Now, integrate each term with respect to \( x \):
\( I = \int \left( \frac{-1/3}{x^2+1} + \frac{4/3}{x^2+4} \right) d x \)
\( I = -\frac{1}{3}\int \frac{1}{x^2+1} d x + \frac{4}{3}\int \frac{1}{x^2+2^2} d x \)
For \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \):
For the first integral, \( a=1 \). For the second integral, \( a=2 \).
\( I = -\frac{1}{3}\tan^{-1}x + \frac{4}{3}\left( \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) \right) + C \)
\( \implies I = -\frac{1}{3}\tan^{-1}x + \frac{2}{3}\tan^{-1}\left(\frac{x}{2}\right) + C \)
In simple words: We used a substitution to change \( x^2 \) to \( t \), which made the fraction easier to handle. Then we broke this new fraction into two simpler parts using partial fractions. After finding the missing numbers, we put \( x^2 \) back and integrated using the arctangent formula for fractions like \( \frac{1}{x^2+a^2} \).

🎯 Exam Tip: When the integrand contains \( x^2 \) terms in a specific pattern, substitution can often simplify the problem significantly. Remember the standard integral forms for \( \int \frac{1}{x^2+a^2} dx \).

Question 19. \( \int \frac{d x}{1+x+x^2+x^3} \)
Answer:
To solve this integral, we first factor the denominator and then use partial fraction decomposition.
Let \( I = \int \frac{d x}{1+x+x^2+x^3} \)
Factor the denominator by grouping terms:
\( 1+x+x^2+x^3 = (1+x) + x^2(1+x) = (1+x)(1+x^2) \)
So, \( I = \int \frac{d x}{(1+x)(1+x^2)} \)
Now, set up the partial fraction decomposition for the integrand. Since \( 1+x^2 \) is an irreducible quadratic factor, its numerator must be linear.
Let \( \frac{1}{(1+x)(1+x^2)} = \frac{\mathrm{A}}{1+x} + \frac{\mathrm{B}x+\mathrm{C}}{1+x^2} \) ...(1)
Multiply both sides of equation (1) by \( (1+x)(1+x^2) \) to clear the denominators:
\( 1 = A(1+x^2) + (Bx+C)(1+x) \) ...(2)
Substitute values for \( x \) and compare coefficients to find A, B, and C.
Substitute \( x = -1 \) into equation (2):
\( 1 = A(1+(-1)^2) + (B(-1)+C)(1-1) \)
\( 1 = A(2) + 0 \)
\( 1 = 2A \)
\( \implies A = \frac{1}{2} \)
Now, expand equation (2):
\( 1 = A+Ax^2 + Bx+Bx^2 + C+Cx \)
\( 1 = (A+B)x^2 + (B+C)x + (A+C) \)
Compare the coefficients of \( x^2 \):
\( 0 = A+B \)
Substitute \( A=\frac{1}{2} \):
\( 0 = \frac{1}{2}+B \)
\( \implies B = -\frac{1}{2} \)
Compare the coefficients of \( x \):
\( 0 = B+C \)
Substitute \( B=-\frac{1}{2} \):
\( 0 = -\frac{1}{2}+C \)
\( \implies C = \frac{1}{2} \)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{1}{(1+x)(1+x^2)} = \frac{1/2}{1+x} + \frac{-1/2 x+1/2}{1+x^2} \)
\( = \frac{1}{2}\left( \frac{1}{1+x} \right) + \frac{1}{2}\left( \frac{-x+1}{1+x^2} \right) \)
Finally, integrate each term:
\( I = \int \frac{1}{2}\left( \frac{1}{1+x} \right) d x + \int \frac{1}{2}\left( \frac{1-x}{1+x^2} \right) d x \)
\( I = \frac{1}{2}\int \frac{1}{1+x} d x + \frac{1}{2}\int \frac{1}{1+x^2} d x - \frac{1}{2}\int \frac{x}{1+x^2} d x \)
\( I = \frac{1}{2}\log|1+x| + \frac{1}{2}\tan^{-1}x - \frac{1}{2}\left( \frac{1}{2}\log|1+x^2| \right) + C \)
\( \implies I = \frac{1}{2}\log|1+x| + \frac{1}{2}\tan^{-1}x - \frac{1}{4}\log(1+x^2) + C \)
In simple words: First, we grouped terms in the denominator to factor it into simple and "unbreakable" parts. Then, we used partial fractions, treating the unbreakable part specially with a linear term on top. We found the missing numbers, and then integrated each part using logarithms and the arctangent formula.

🎯 Exam Tip: Remember to factor the denominator completely first. If you encounter an irreducible quadratic factor (like \( 1+x^2 \)), its corresponding partial fraction term will have a linear numerator, \( \frac{Bx+C}{Ax^2+Bx+C} \).

Question 20. \( \int \frac{x^2 d x}{\left(x^2-1\right)\left(x^2+2\right)} \)
Answer:
To solve this integral, we first use a substitution to simplify the integrand.
Let \( I = \int \frac{x^2 d x}{\left(x^2-1\right)\left(x^2+2\right)} \)
Let \( t = x^2 \)
Then the integrand becomes \( \frac{t}{(t-1)(t+2)} \)
Now, set up the partial fraction decomposition for the new integrand:
Let \( \frac{t}{(t-1)(t+2)} = \frac{\mathrm{A}}{t-1} + \frac{\mathrm{B}}{t+2} \) ...(1)
Multiply both sides of equation (1) by \( (t-1)(t+2) \) to clear the denominators:
\( t = A(t+2) + B(t-1) \) ...(2)
Substitute values for \( t \) to find A and B.
Substitute \( t = 1 \) into equation (2):
\( 1 = A(1+2) + B(1-1) \)
\( 1 = A(3) + B(0) \)
\( 1 = 3A \)
\( \implies A = \frac{1}{3} \)
Substitute \( t = -2 \) into equation (2):
\( -2 = A(-2+2) + B(-2-1) \)
\( -2 = A(0) + B(-3) \)
\( -2 = -3B \)
\( \implies B = \frac{2}{3} \)
Now, substitute the values of A and B back into equation (1):
\( \frac{t}{(t-1)(t+2)} = \frac{1/3}{t-1} + \frac{2/3}{t+2} \)
Substitute back \( t = x^2 \):
\( \frac{x^2}{(x^2-1)(x^2+2)} = \frac{1/3}{x^2-1} + \frac{2/3}{x^2+2} \)
Now, integrate each term with respect to \( x \):
\( I = \int \left( \frac{1/3}{x^2-1} + \frac{2/3}{x^2+2} \right) d x \)
\( I = \frac{1}{3}\int \frac{1}{x^2-1} d x + \frac{2}{3}\int \frac{1}{x^2+(\sqrt{2})^2} d x \)
For \( \int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| \), use \( a=1 \).
For \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \), use \( a=\sqrt{2} \).
\( I = \frac{1}{3}\left( \frac{1}{2(1)}\log\left|\frac{x-1}{x+1}\right| \right) + \frac{2}{3}\left( \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) \right) + C \)
\( \implies I = \frac{1}{6}\log\left|\frac{x-1}{x+1}\right| + \frac{\sqrt{2}}{3}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + C \)
In simple words: We made the integral simpler by changing \( x^2 \) to \( t \). This helped us break the fraction into two simpler parts using partial fractions. After finding the missing numbers for these parts, we put \( x^2 \) back and integrated using special formulas for fractions like \( \frac{1}{x^2-a^2} \) and \( \frac{1}{x^2+a^2} \).

🎯 Exam Tip: Remember to express the denominators in the form \( x^2-a^2 \) or \( x^2+a^2 \) after substitution to correctly apply the standard integral formulas.

Question 21. \( \int \frac{2 x}{x^3-1} d x \)
Answer:
To solve this integral, we first factor the denominator and then use partial fraction decomposition.
Let \( I = \int \frac{2 x}{x^3-1} d x \)
Factor the denominator using the difference of cubes formula \( a^3-b^3 = (a-b)(a^2+ab+b^2) \):
\( x^3-1 = (x-1)(x^2+x+1) \)
So, \( I = \int \frac{2 x}{(x-1)(x^2+x+1)} d x \)
Now, set up the partial fraction decomposition for the integrand. Since \( x^2+x+1 \) is an irreducible quadratic factor, its numerator must be linear.
Let \( \frac{2 x}{(x-1)(x^2+x+1)} = \frac{\mathrm{A}}{x-1} + \frac{\mathrm{B}x+\mathrm{C}}{x^2+x+1} \) ...(1)
Multiply both sides of equation (1) by \( (x-1)(x^2+x+1) \) to clear the denominators:
\( 2x = A(x^2+x+1) + (Bx+C)(x-1) \) ...(2)
Substitute values for \( x \) and compare coefficients to find A, B, and C.
Substitute \( x = 1 \) into equation (2):
\( 2(1) = A(1^2+1+1) + (B(1)+C)(1-1) \)
\( 2 = A(3) + 0 \)
\( 2 = 3A \)
\( \implies A = \frac{2}{3} \)
Now, expand equation (2):
\( 2x = Ax^2+Ax+A + Bx^2-Bx+Cx-C \)
\( 2x = (A+B)x^2 + (A-B+C)x + (A-C) \)
Compare the coefficients of \( x^2 \):
\( 0 = A+B \)
Substitute \( A=\frac{2}{3} \):
\( 0 = \frac{2}{3}+B \)
\( \implies B = -\frac{2}{3} \)
Compare the coefficients of the constant term (terms without \( x \)):
\( 0 = A-C \)
Substitute \( A=\frac{2}{3} \):
\( 0 = \frac{2}{3}-C \)
\( \implies C = \frac{2}{3} \)
(We can also check the coefficient of \( x \): \( A-B+C = \frac{2}{3} - (-\frac{2}{3}) + \frac{2}{3} = \frac{2}{3}+\frac{2}{3}+\frac{2}{3} = \frac{6}{3} = 2 \). This matches the left side \( 2x \), confirming our values.)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{2x}{(x-1)(x^2+x+1)} = \frac{2/3}{x-1} + \frac{-2/3 x+2/3}{x^2+x+1} \)
\( = \frac{2}{3}\left( \frac{1}{x-1} \right) + \frac{2}{3}\left( \frac{-x+1}{x^2+x+1} \right) \)
Finally, integrate each term:
\( I = \int \frac{2}{3}\left( \frac{1}{x-1} \right) d x + \int \frac{2}{3}\left( \frac{1-x}{x^2+x+1} \right) d x \)
\( I = \frac{2}{3}\log|x-1| + \frac{2}{3}\int \frac{1-x}{x^2+x+1} d x \)
For the second integral, we adjust the numerator to match the derivative of the denominator and a constant term.
Derivative of \( x^2+x+1 \) is \( 2x+1 \). We need \( 1-x \).
\( \frac{1-x}{x^2+x+1} = -\frac{x-1}{x^2+x+1} = -\frac{1}{2}\frac{2x-2}{x^2+x+1} = -\frac{1}{2}\frac{2x+1-3}{x^2+x+1} = -\frac{1}{2}\left( \frac{2x+1}{x^2+x+1} - \frac{3}{x^2+x+1} \right) \)
So, \( \int \frac{1-x}{x^2+x+1} d x = -\frac{1}{2}\int \frac{2x+1}{x^2+x+1} d x + \frac{3}{2}\int \frac{1}{x^2+x+1} d x \)
\( = -\frac{1}{2}\log|x^2+x+1| + \frac{3}{2}\int \frac{1}{(x+1/2)^2 + (\sqrt{3}/2)^2} d x \)
\( = -\frac{1}{2}\log|x^2+x+1| + \frac{3}{2}\left( \frac{1}{\sqrt{3}/2}\tan^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}\right) \right) \)
\( = -\frac{1}{2}\log|x^2+x+1| + \frac{3}{2}\left( \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) \right) \)
\( = -\frac{1}{2}\log|x^2+x+1| + \sqrt{3}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) \)
Now, combine all parts for \( I \):
\( I = \frac{2}{3}\log|x-1| + \frac{2}{3}\left( -\frac{1}{2}\log|x^2+x+1| + \sqrt{3}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) \right) + C \)
\( \implies I = \frac{2}{3}\log|x-1| - \frac{1}{3}\log|x^2+x+1| + \frac{2\sqrt{3}}{3}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C \)
In simple words: We started by breaking the denominator using the difference of cubes rule. Then, we used partial fractions, knowing that the quadratic part needed a linear term on top. We found the missing numbers by comparing coefficients and substituting values. Finally, we integrated each part, carefully handling the quadratic term by completing the square and using the arctangent formula.

🎯 Exam Tip: When dealing with irreducible quadratic factors, remember to complete the square in the denominator to apply the \( \tan^{-1} \) integration formula. Also, carefully adjust the numerator to separate terms for \( \log \) and \( \tan^{-1} \) integration.

Question 22. \( \int \frac{d x}{x+x^2+x^3} \)
Answer:
To solve this integral, we first factor the denominator and then use partial fraction decomposition.
Let \( I = \int \frac{d x}{x+x^2+x^3} \)
Factor the denominator by taking \( x \) common and then grouping:
\( x+x^2+x^3 = x(1+x+x^2) \)
So, \( I = \int \frac{d x}{x(1+x+x^2)} \)
Now, set up the partial fraction decomposition for the integrand. Since \( 1+x+x^2 \) is an irreducible quadratic factor, its numerator must be linear.
Let \( \frac{1}{x(1+x+x^2)} = \frac{\mathrm{A}}{x} + \frac{\mathrm{B}x+\mathrm{C}}{1+x+x^2} \) ...(1)
Multiply both sides of equation (1) by \( x(1+x+x^2) \) to clear the denominators:
\( 1 = A(1+x+x^2) + (Bx+C)x \) ...(2)
Substitute values for \( x \) and compare coefficients to find A, B, and C.
Substitute \( x = 0 \) into equation (2):
\( 1 = A(1+0+0) + (B(0)+C)(0) \)
\( 1 = A(1) + 0 \)
\( \implies A = 1 \)
Now, expand equation (2):
\( 1 = A+Ax+Ax^2 + Bx^2+Cx \)
\( 1 = (A+B)x^2 + (A+C)x + A \)
Compare the coefficients of \( x^2 \):
\( 0 = A+B \)
Substitute \( A=1 \):
\( 0 = 1+B \)
\( \implies B = -1 \)
Compare the coefficients of \( x \):
\( 0 = A+C \)
Substitute \( A=1 \):
\( 0 = 1+C \)
\( \implies C = -1 \)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{1}{x(1+x+x^2)} = \frac{1}{x} + \frac{-x-1}{1+x+x^2} \)
\( = \frac{1}{x} - \frac{x+1}{1+x+x^2} \)
Finally, integrate each term:
\( I = \int \frac{1}{x} d x - \int \frac{x+1}{1+x+x^2} d x \)
\( I = \log|x| - \int \frac{x+1}{x^2+x+1} d x \)
For the second integral, we adjust the numerator to match the derivative of the denominator \( (2x+1) \) and a constant term.
\( \frac{x+1}{x^2+x+1} = \frac{1}{2}\frac{2x+2}{x^2+x+1} = \frac{1}{2}\frac{2x+1+1}{x^2+x+1} = \frac{1}{2}\left( \frac{2x+1}{x^2+x+1} + \frac{1}{x^2+x+1} \right) \)
So, \( \int \frac{x+1}{x^2+x+1} d x = \frac{1}{2}\int \frac{2x+1}{x^2+x+1} d x + \frac{1}{2}\int \frac{1}{x^2+x+1} d x \)
\( = \frac{1}{2}\log|x^2+x+1| + \frac{1}{2}\int \frac{1}{(x+1/2)^2 + (\sqrt{3}/2)^2} d x \)
\( = \frac{1}{2}\log|x^2+x+1| + \frac{1}{2}\left( \frac{1}{\sqrt{3}/2}\tan^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}\right) \right) \)
\( = \frac{1}{2}\log|x^2+x+1| + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) \)
Now, combine all parts for \( I \):
\( I = \log|x| - \left( \frac{1}{2}\log|x^2+x+1| + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) \right) + C \)
\( \implies I = \log|x| - \frac{1}{2}\log|x^2+x+1| - \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C \)
In simple words: We first factored the denominator by taking out \( x \), leaving a linear term and an unbreakable quadratic term. Then, we used partial fractions to split the original fraction into two parts, one for each factor. We found the missing numbers and integrated each part separately, using logarithms for the linear term and a combination of logarithm and arctangent for the quadratic part after completing the square.

🎯 Exam Tip: Always look for common factors in the denominator before trying other factorization methods. For irreducible quadratic factors, completing the square is essential for applying the arctangent integration formula.

Question 23. \( \int \frac{\tan \theta+\tan ^3 \theta}{1+\tan ^3 \theta} d \theta \)
Answer:
To solve this integral, we use a substitution to simplify the expression, then apply partial fraction decomposition.
Let \( I = \int \frac{\tan \theta+\tan ^3 \theta}{1+\tan ^3 \theta} d \theta \)
Factor the numerator: \( \tan \theta+\tan ^3 \theta = \tan \theta(1+\tan^2\theta) = \tan \theta \sec^2\theta \)
So, \( I = \int \frac{\tan \theta \sec^2 \theta}{1+\tan ^3 \theta} d \theta \)
Let \( t = \tan \theta \)
Then, \( d t = \sec^2 \theta d \theta \)
Substitute \( t \) and \( dt \) into the integral:
\( I = \int \frac{t d t}{1+t^3} \)
Factor the denominator using the sum of cubes formula \( a^3+b^3 = (a+b)(a^2-ab+b^2) \):
\( 1+t^3 = (1+t)(1-t+t^2) \)
So, \( I = \int \frac{t d t}{(1+t)(1-t+t^2)} \)
Now, set up the partial fraction decomposition. Since \( 1-t+t^2 \) is an irreducible quadratic factor, its numerator must be linear.
Let \( \frac{t}{(1+t)(1-t+t^2)} = \frac{\mathrm{A}}{1+t} + \frac{\mathrm{B}t+\mathrm{C}}{1-t+t^2} \) ...(1)
Multiply both sides of equation (1) by \( (1+t)(1-t+t^2) \) to clear the denominators:
\( t = A(1-t+t^2) + (Bt+C)(1+t) \) ...(2)
Substitute values for \( t \) and compare coefficients to find A, B, and C.
Substitute \( t = -1 \) into equation (2):
\( -1 = A(1-(-1)+(-1)^2) + (B(-1)+C)(1-1) \)
\( -1 = A(1+1+1) + 0 \)
\( -1 = 3A \)
\( \implies A = -\frac{1}{3} \)
Now, expand equation (2):
\( t = A-At+At^2 + Bt+Bt^2 + C+Ct \)
\( t = (A+B)t^2 + (-A+B+C)t + (A+C) \)
Compare the coefficients of \( t^2 \):
\( 0 = A+B \)
Substitute \( A=-\frac{1}{3} \):
\( 0 = -\frac{1}{3}+B \)
\( \implies B = \frac{1}{3} \)
Compare the coefficients of \( t \):
\( 1 = -A+B+C \)
Substitute \( A=-\frac{1}{3} \) and \( B=\frac{1}{3} \):
\( 1 = -(-\frac{1}{3}) + \frac{1}{3} + C \)
\( 1 = \frac{1}{3} + \frac{1}{3} + C \)
\( 1 = \frac{2}{3} + C \)
\( \implies C = 1 - \frac{2}{3} = \frac{1}{3} \)
Now, substitute the values of A, B, and C back into equation (1):
\( \frac{t}{(1+t)(1-t+t^2)} = \frac{-1/3}{1+t} + \frac{1/3 t+1/3}{1-t+t^2} \)
\( = -\frac{1}{3}\left( \frac{1}{1+t} \right) + \frac{1}{3}\left( \frac{t+1}{t^2-t+1} \right) \)
Finally, integrate each term:
\( I = -\frac{1}{3}\int \frac{1}{1+t} d t + \frac{1}{3}\int \frac{t+1}{t^2-t+1} d t \)
\( I = -\frac{1}{3}\log|1+t| + \frac{1}{3}\int \frac{t+1}{t^2-t+1} d t \)
For the second integral, we adjust the numerator to match the derivative of the denominator \( (2t-1) \) and a constant term.
\( \frac{t+1}{t^2-t+1} = \frac{1}{2}\frac{2t+2}{t^2-t+1} = \frac{1}{2}\frac{2t-1+3}{t^2-t+1} = \frac{1}{2}\left( \frac{2t-1}{t^2-t+1} + \frac{3}{t^2-t+1} \right) \)
So, \( \int \frac{t+1}{t^2-t+1} d t = \frac{1}{2}\int \frac{2t-1}{t^2-t+1} d t + \frac{3}{2}\int \frac{1}{t^2-t+1} d t \)
\( = \frac{1}{2}\log|t^2-t+1| + \frac{3}{2}\int \frac{1}{(t-1/2)^2 + (\sqrt{3}/2)^2} d t \)
\( = \frac{1}{2}\log|t^2-t+1| + \frac{3}{2}\left( \frac{1}{\sqrt{3}/2}\tan^{-1}\left(\frac{t-1/2}{\sqrt{3}/2}\right) \right) \)
\( = \frac{1}{2}\log|t^2-t+1| + \frac{3}{2}\left( \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right) \right) \)
\( = \frac{1}{2}\log|t^2-t+1| + \sqrt{3}\tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right) \)
Now, combine all parts for \( I \):
\( I = -\frac{1}{3}\log|1+t| + \frac{1}{3}\left( \frac{1}{2}\log|t^2-t+1| + \sqrt{3}\tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right) \right) + C \)
\( I = -\frac{1}{3}\log|1+t| + \frac{1}{6}\log|t^2-t+1| + \frac{\sqrt{3}}{3}\tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right) + C \)
Substitute back \( t = \tan \theta \):
\( \implies I = -\frac{1}{3}\log|1+\tan \theta| + \frac{1}{6}\log|\tan^2 \theta-\tan \theta+1| + \frac{\sqrt{3}}{3}\tan^{-1}\left(\frac{2\tan \theta-1}{\sqrt{3}}\right) + C \)
In simple words: We first simplified the integral by replacing \( \tan \theta \) with \( t \). Then we factored the denominator and used partial fractions, making sure to handle the 'unbreakable' quadratic part with a linear numerator. We found the missing numbers and then integrated each piece. For the quadratic part, we had to complete the square to use the arctangent formula. Finally, we changed \( t \) back to \( \tan \theta \).

🎯 Exam Tip: For integrals involving \( \tan \theta \) and \( \sec^2 \theta \), the substitution \( t=\tan \theta \) is very common and effective. Always remember to factor cubic denominators for partial fractions and handle irreducible quadratic factors correctly by completing the square.

Question 24. \( \int \frac{\sec ^2 \theta d \theta}{\tan ^3 \theta+4 \tan \theta} \)
Answer: Let the integral be \( I \).
First, we substitute \( t = \tan \theta \).
Then, \( d t = \sec^2 \theta d \theta \).
Now, the integral becomes:
\( I = \int \frac{d t}{t^3+4t} \)
We can factor out \( t \) from the denominator:
\( I = \int \frac{d t}{t(t^2+4)} \)
This is a partial fraction problem. We express the fraction as:
\( \frac{1}{t(t^2+4)} = \frac{\mathrm{A}}{t} + \frac{\mathrm{B}t+\mathrm{C}}{t^2+4} \)
Multiply both sides by \( t(t^2+4) \):
\( 1 = \mathrm{A}(t^2+4) + (\mathrm{B}t+\mathrm{C})t \)
When \( t=0 \):
\( 1 = \mathrm{A}(0^2+4) + (\mathrm{B}(0)+\mathrm{C})(0) \)
\( 1 = 4\mathrm{A} \)
\( \implies \) \( \mathrm{A} = \frac{1}{4} \)
Compare coefficients of \( t^2 \):
\( 0 = \mathrm{A} + \mathrm{B} \)
\( \implies \) \( 0 = \frac{1}{4} + \mathrm{B} \)
\( \implies \) \( \mathrm{B} = -\frac{1}{4} \)
Compare coefficients of \( t \):
\( 0 = \mathrm{C} \)
So, the partial fraction decomposition is:
\( \frac{1}{t(t^2+4)} = \frac{1/4}{t} + \frac{(-1/4)t}{t^2+4} \)
Now, we integrate:
\( I = \int \left( \frac{1/4}{t} - \frac{t/4}{t^2+4} \right) d t \)
\( I = \frac{1}{4}\int \frac{1}{t} d t - \frac{1}{4}\int \frac{t}{t^2+4} d t \)
For the second integral, we can multiply and divide by 2 to make the numerator the derivative of the denominator:
\( I = \frac{1}{4}\int \frac{1}{t} d t - \frac{1}{8}\int \frac{2t}{t^2+4} d t \)
\( I = \frac{1}{4}\log|t| - \frac{1}{8}\log|t^2+4| + C \)
We can combine the logarithms:
\( I = \frac{1}{8} \left( 2\log|t| - \log|t^2+4| \right) + C \)
\( I = \frac{1}{8} \left( \log|t^2| - \log|t^2+4| \right) + C \)
\( I = \frac{1}{8} \log\left|\frac{t^2}{t^2+4}\right| + C \)
Finally, substitute back \( t = \tan \theta \):
\( I = \frac{1}{8} \log\left|\frac{\tan^2 \theta}{\tan^2 \theta+4}\right| + C \)
This method of partial fractions helps simplify complex integrals into basic forms. Remember that \( \frac{1}{4} \log\left|\frac{\tan \theta}{\sqrt{\tan^2 \theta+4}}\right| \) is an equivalent form by taking the square root out of the logarithm.In simple words: We first change \( \tan \theta \) to \( t \) to make the integral simpler. Then we break the fraction into smaller, easier-to-integrate parts using a method called partial fractions. After integrating each part, we put \( \tan \theta \) back into the answer.

🎯 Exam Tip: When using substitution for integrals involving trigonometric functions, always remember to change the differential term (\( d \theta \)) as well as the function itself. Always convert back to the original variable at the end.

Question 25. \( \int \frac{d x}{\sin x+\tan x} \)
Answer: Let the integral be \( I \).
First, rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \):
\( I = \int \frac{d x}{\sin x+\frac{\sin x}{\cos x}} \)
Combine the terms in the denominator:
\( I = \int \frac{d x}{\frac{\sin x \cos x + \sin x}{\cos x}} \)
Simplify the fraction:
\( I = \int \frac{\cos x d x}{\sin x (\cos x+1)} \)
This integral can be solved using substitution. Let \( u = \cos x \).
Then, \( d u = -\sin x d x \), so \( \sin x d x = -d u \).
Also, use the identity \( \sin^2 x = 1-\cos^2 x \).
We can rewrite the integrand by adding and subtracting 1 in the numerator:
\( I = \int \frac{1+\cos x - 1}{\sin x(1+\cos x)} d x \)
\( I = \int \left( \frac{1}{\sin x} - \frac{1}{\sin x(1+\cos x)} \right) d x \)
\( I = \int \csc x d x - \int \frac{1}{\sin x(1+\cos x)} d x \)
The first part is a standard integral: \( \int \csc x d x = \log|\csc x - \cot x| \).
For the second part, let's call it \( I_1 \):
\( I_1 = \int \frac{1}{\sin x(1+\cos x)} d x \)
Rewrite \( \sin x \) as \( \frac{1-\cos^2 x}{\sin x} = \frac{(1-\cos x)(1+\cos x)}{\sin x} \). This is tricky. Let's use the substitution \( t = \cos x \) more directly here.
Then \( d t = -\sin x d x \). So \( d x = \frac{-d t}{\sin x} \).
\( I_1 = \int \frac{1}{\sin x (1+t)} \frac{-d t}{\sin x} = \int \frac{-d t}{\sin^2 x (1+t)} \)
Since \( \sin^2 x = 1-\cos^2 x = 1-t^2 \), we get:
\( I_1 = \int \frac{-d t}{(1-t^2)(1+t)} \)
Factor the denominator:
\( I_1 = \int \frac{-d t}{(1-t)(1+t)(1+t)} = \int \frac{-d t}{(1-t)(1+t)^2} \)
Now, use partial fractions for \( \frac{1}{(1-t)(1+t)^2} \):
\( \frac{1}{(1-t)(1+t)^2} = \frac{\mathrm{A}}{1-t} + \frac{\mathrm{B}}{1+t} + \frac{\mathrm{C}}{(1+t)^2} \)
Multiply by \( (1-t)(1+t)^2 \):
\( 1 = \mathrm{A}(1+t)^2 + \mathrm{B}(1-t)(1+t) + \mathrm{C}(1-t) \)
Set \( t=1 \):
\( 1 = \mathrm{A}(1+1)^2 \)
\( 1 = 4\mathrm{A} \)
\( \implies \) \( \mathrm{A} = \frac{1}{4} \)
Set \( t=-1 \):
\( 1 = \mathrm{C}(1-(-1)) \)
\( 1 = 2\mathrm{C} \)
\( \implies \) \( \mathrm{C} = \frac{1}{2} \)
Compare coefficients of \( t^2 \):
\( 0 = \mathrm{A} - \mathrm{B} \)
\( \implies \) \( 0 = \frac{1}{4} - \mathrm{B} \)
\( \implies \) \( \mathrm{B} = \frac{1}{4} \)
So, \( \frac{1}{(1-t)(1+t)^2} = \frac{1/4}{1-t} + \frac{1/4}{1+t} + \frac{1/2}{(1+t)^2} \).
Now, integrate \( I_1 \) (remembering the negative sign from the substitution):
\( I_1 = -\int \left( \frac{1/4}{1-t} + \frac{1/4}{1+t} + \frac{1/2}{(1+t)^2} \right) d t \)
\( I_1 = -\frac{1}{4}\int \frac{1}{1-t} d t - \frac{1}{4}\int \frac{1}{1+t} d t - \frac{1}{2}\int (1+t)^{-2} d t \)
\( I_1 = -\frac{1}{4} (-\log|1-t|) - \frac{1}{4}\log|1+t| - \frac{1}{2} \frac{(1+t)^{-1}}{-1} + C \)
\( I_1 = \frac{1}{4}\log|1-t| - \frac{1}{4}\log|1+t| + \frac{1}{2(1+t)} + C \)
Combine the logarithmic terms:
\( I_1 = \frac{1}{4}\log\left|\frac{1-t}{1+t}\right| + \frac{1}{2(1+t)} + C \)
Substitute back \( t = \cos x \):
\( I_1 = \frac{1}{4}\log\left|\frac{1-\cos x}{1+\cos x}\right| + \frac{1}{2(1+\cos x)} + C \)
So the final integral \( I \) is:
\( I = \log|\csc x - \cot x| + \frac{1}{4}\log\left|\frac{1-\cos x}{1+\cos x}\right| + \frac{1}{2(1+\cos x)} + C \)
This approach helps integrate expressions that cannot be solved by direct substitution.In simple words: First, we change \( \tan x \) to \( \sin x / \cos x \) to get a common denominator. Then, we use a substitution, letting \( t = \cos x \), which changes the integral into a form that can be solved using partial fractions. After splitting the fraction into simpler parts and integrating each, we put \( \cos x \) back in place of \( t \) to get the final answer.

🎯 Exam Tip: For complex trigonometric integrals, converting all terms to \( \sin x \) and \( \cos x \) is often the first step. Look for opportunities to use substitutions like \( t = \cos x \) or \( t = \sin x \) and then apply partial fractions if the resulting algebraic expression is suitable.